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Tomtit [17]
3 years ago
12

True or false if the dimensions of a cylinder are doubled then its volume is quadrupled.

Mathematics
2 answers:
IgorC [24]3 years ago
8 0
VOLUME OF A CYLINDER = pi X r X r X h 

<span>pi X 2r X 2r X 2h = 8pir^2h </span>

<span>Volume is 8 times </span>

<span>ANSWER IS FALSE</span>
Travka [436]3 years ago
7 0

Answer:  The given statement is FALSE.

Step-by-step explanation:  We are given to check whether the following statement is true or false :

"If the dimensions of a cylinder are doubled then its volume is quadrupled."

We know that

the volume of a cylinder with radius r units and height h units is given by

V=\pi r^2h.

If the dimensions are doubled, then the new dimensions will be

R = 2r  and  H = 2h.

Therefore, the new volume will be

V'=\pi R^2H=\pi \times(2r)^2\times (2h)=8\times \pi r^2 h=8V.

Thus, the volume of the new cylinder will be 8 times the original one, not quadrupled.

So, the given statement is FALSE.

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Farah's brother is 7 years less than twice Farah's age, f. The sum of Farah's age and
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f+(2f-7)=41

Step-by-step explanation:

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Which ordered pair is a solution of the equation?
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7 0
3 years ago
Find the solutions of the quadratic equation 3x^2-5x+1=0.
Black_prince [1.1K]

Answer:

The solutions of the quadratic equation are x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}

Step-by-step explanation:

This is a second order polynomial, and we can find it's roots by the Bhaskara formula.

Explanation of the bhaskara formula:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

For this problem, we have to find x_{1} \text{and} x_{2}.

The polynomial is 3x^{2} - 5x +1, so a = 3, b = -5, c = 1.

Solution

\bigtriangleup = b^{2} - 4ac = (-5)^{2} - 4*3*1 = 25 - 12 = 13

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) + \sqrt{13}}{2*3} = \frac{5 + \sqrt{13}}{6}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) - \sqrt{13}}{2*3} = \frac{5 - \sqrt{13}}{6}

The solutions of the quadratic equation are x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}

3 0
3 years ago
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